Network Summarisation – Exam Technique and Examples

Problem

Note: Yes I'm spelling Summarisation with an â€™Sâ€™ Iâ€™m English.

Most examples I've seen on this give you a bunch of subnets then ask you to come up with a summary route for all of them, (that's kind of the point of route summarisation, I'll grant you). However in an exam with a laminated board and the dodgy permanent pen they give you to make notes with, are you seriously expected to convert everything to binary to find the â€˜last contigious bitâ€™?

I was studying this today and kept getting it wrong, so I askedÂ a colleague to look andÂ see where I was going wrong. He scribbled on a piece of paper for two minutes and came up with the correct answer. So Iâ€™ve stolen his methodology. You can use it for any range of subnets, and if it comes up in an exam, you also have the added bonus that the right answer will be on the screen so even if your maths is off a little, the answer should jump out at you.

Step 2: Write down the Highest and Lowest Network (in fact just the third octet).

10
15

Step 3:Convert those into Binary (use a full 8 bits).

10 = 00001010 (if youâ€™ve just gone eh! That's an 8, and a 2).
15 = 00001111 (if youâ€™ve just gone eh! That's an 8, a 4, a 2, and a 1).

Step 4:Â Find the POSITION of last contigious bit, (the same in both).

00001010
00001111

Above, the first 5 numbers are the same, thats where the LAST bit of the summarised subnet will be. (If you're confused, everything to the left will be a one, everything to the right will be a zero. i.e. 11111111.11111111.11111000.0000000

So the answer will have a /21 mask, (because thereâ€™s 21 x 1s).

Well that's great, but I still donâ€™t know the subnet address?

Yes you do! In step 3 you worked out the LOWEST subnet, you simply forget all the numbers that are NOT contigious, and the subnet is all the bits that are left. (That sounds more complicated than it is), So;

0000101000001111

You ignore the last three, they are not contigious, (010 and 111 are not the same), that leaves you with;

00001000 = 8

So the network to summarise is 192.168.8.0/21 (or 255.255.248.0 if you prefer).

You might think, THAT WAS LONG WINDED! Well I took pains to explain everything. Once you have the method you can apply it to any list of networks;.

A More Complicated Example

192.168.112.0 /24
192.168.113.0 /24
192.168.114.0 /24
192.168.115.0 /24
192.168.116.0 /24
192.168.117.0 /24
192.168.118.0 /24
192.168.119.0 /24
Lowest and Higest
192.168.112.0
192.168.119.0
Convert the changing Octet to Binary
01110000
01110111
We are the same up to the 5th bit so its a /21Drop the last three bits 01110000 is 112
Answer is 192.168.112.0/21